Borel Measure is Outer Regular iff it is inner regular?

I have a doubt on whether this theorem is true: Let $(X,\Gamma)$ be a compact space. Then a Borel measure $\mu$ on $\mathbb{B}(X)$ is outer regular iff it is inner regular. Can anyone shed some light in helping me to prove this? After reading through various books, I am thinking that proving this is possible with $\mu(x) < +\infty$ but am not certain.

• So, what can you do in case $\mu(X) < \infty$? – GEdgar Oct 22 '13 at 17:32
• I am not sure, GEdgar. I found this information from Bauer's book under Measures on Topological Spaces and am still thinking. – Sandra Oct 22 '13 at 19:07
• Anyone can shed some light on this question? – Sandra Oct 23 '13 at 17:03
• You need to do some work. Write down the definition of inner regular. Write down the definition of outer regular. Investigate how approximating a set $A$ from inside is related to approximating the complement $X \setminus A$ from outside. – GEdgar Oct 23 '13 at 17:23
• @GEdgar, I did! Writing down the definition of outer regular and inner regular is the basic starting of the proof. The question I am seeking is the conclusion part of the theorem "iff it is inner regular". I am well aware too that each Borel subset of the compact space satisfy both definition but from my understanding of this theorem, we need to show it is inner regular first then only outer regular which I am not sure I am right and am lost. – Sandra Oct 23 '13 at 17:29